Quantum physics is an HL-only topic about the particle behaviour of light and the wave behaviour of matter. The photoelectric effect, with work function, threshold frequency and stopping potential, shows that light transfers energy in photons, and the graph of stopping potential against frequency gives the Planck constant.
Matter waves, λ = h/p, are tested through electron and neutron diffraction, and Compton scattering shows that photons carry momentum. Questions also ask why intensity changes the photoelectric current but not the maximum kinetic energy of the electrons.
45 questions
210 marks
Paper 1A: 27
Paper 1B: 7
Paper 2: 11
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21 practice questions on E.2 Quantum physics
1E-1A-04
The photoelectric effect·E.2 Quantum physics (HL)
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
Light of frequency f ejects photoelectrons of maximum kinetic energy Ek from a metal of work function Φ. The frequency of the light is doubled.
What is the new maximum kinetic energy?
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Notes
Step 1Einstein's equation: Ek = hf − Φ, so hf = Ek + Φ.
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All 3 steps must be completed — there is no mark for a part-answer.
AKinetic energy is not proportional to frequency because of the fixed work function.
BThis uses 2Ek in place of 2hf: 2Ek − Φ.
CCorrect: doubling the photon energy adds Ek + Φ to the original kinetic energy.
DThis is twice the photon energy — the work function still has to be paid once.
Syllabus understandingE.2 (HL) — the photoelectric effect as evidence of the particle nature of light; Einstein's explanation using the work function and the maximum kinetic energy of the photoelectrons as given by Emax = hf − Φ Command term: Deduce
2E-1A-05
Matter waves·E.2 Quantum physics (HL)
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
An electron and a proton have the same kinetic energy.
What is the ratio (de Broglie wavelength of the electron) / (de Broglie wavelength of the proton)?
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Notes
Step 1λ = h/p, and for a non-relativistic particle p = √(2mEk).
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2At equal kinetic energy λ ∝ 1/√m.
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Step 3λe/λp = √(mp/me) ≈ 43: the lighter electron has the longer wavelength.
✓ 1
Answer D
Answer: D · 3 stages of work, one mark
Every option, and why
AThis would make the electron wavelength shorter, and omits the square root.
BThis omits the square root.
CThis is the inverse — the electron, being lighter, has the longer wavelength.
DCorrect: λ = h/√(2mEk) ∝ 1/√m.
Syllabus understandingE.2 (HL) — that matter exhibits wave–particle duality; the de Broglie wavelength for particles as given by λ = h/p; A.3 — Ek = p²/2mCommand term: Deduce
3E-1A-06
Compton scattering·E.2 Quantum physics (HL)
Paper 1AHard1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
X-ray photons of wavelength 70 pm and gamma-ray photons of wavelength 2.0 pm are each scattered through 90° by free electrons that are initially at rest.
Which row compares the shift in wavelength of the scattered photons and the fraction of the incident photon energy that is transferred to the electron?
Shift in wavelengthFraction of the photon energy transferred to the electron
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Notes
Step 1Δλ = (h/mec)(1 − cos θ) depends only on the angle, not on the incident wavelength: at 90° both shifts equal λC = 2.43 pm.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2The photon energy is hc/λ, so the fraction transferred is 1 − λ/(λ + Δλ) = Δλ/(λ + Δλ).
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Step 3X-ray: 2.43/(70 + 2.43) = 0.033 (about 3 %); gamma ray: 2.43/(2.0 + 2.43) = 0.55 (about 55 %). The same shift is a much larger fraction of a short wavelength.
✓ 1
Answer A
Answer: A · 3 stages of work, one mark
Every option, and why
ACorrect: the shift is fixed by the angle alone, and the same shift is a far larger fraction of the shorter (gamma-ray) wavelength, so that photon loses a larger fraction of its energy.
BThe shift is right, but an equal shift in wavelength does not mean an equal fraction of the energy: the fraction is Δλ/(λ + Δλ), which depends on λ.
CThis takes the more energetic photon to be shifted further. The Compton shift does not depend on the incident wavelength or energy.
DThis takes the shift to be proportional to the incident wavelength (a fixed fractional shift), which would make the fraction of energy transferred the same for both.
Syllabus understandingE.2 (HL) — that photons scatter off electrons with increased wavelength; the shift in photon wavelength after scattering off an electron as given by λf − λi = Δλ = (h/mec)(1 − cos θ); E.1 — E = hf for a photon Command term: Deduce
4E-1A-16
Compton scattering·E.2 Quantum physics (HL)
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDetermine
An X-ray photon of wavelength 10.0 pm is scattered by a free electron. The Compton wavelength of the electron is 2.43 pm.
What is the largest possible wavelength of the scattered photon?
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Notes
Step 1The Compton shift is Δλ = λC(1 − cos θ), where λC = h/mec = 2.43 pm.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2The shift is greatest for back-scattering, θ = 180°, where 1 − cos θ = 2.
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Step 3λ′ = 10.0 + 2 × 2.43 = 14.9 pm.
✓ 1
Answer D
Answer: D · 3 stages of work, one mark
Every option, and why
AThe scattered photon loses energy, so its wavelength must increase.
BNo shift corresponds to θ = 0 — forward scattering, the smallest wavelength, not the largest.
CThis is the shift for θ = 90° (one Compton wavelength).
DCorrect: maximum shift 2λC = 4.86 pm.
Syllabus understandingE.2 (HL) — that photons scatter off electrons with increased wavelength; the shift in photon wavelength after scattering off an electron as given by λf − λi = Δλ = (h/mec)(1 − cos θ) Command term: Determine
5E-1A-18
Matter waves·E.2 Quantum physics (HL)
Paper 1AHard1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
Electrons, accelerated from rest through a potential difference V, pass through a narrow slit. The first minimum of the diffraction pattern is at a small angle θ to the central direction.
The potential difference is reduced to V/4. What is the new angle of the first minimum?
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Notes
Step 1The kinetic energy eV = p²/2m, so p = √(2meV) ∝ √V.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2λ = h/p ∝ 1/√V: dividing V by 4 doubles λ.
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Step 3For a single slit θ ≈ λ/b, so the angle doubles to 2θ.
✓ 1
Answer C
Answer: C · 3 stages of work, one mark
Every option, and why
AThis takes λ ∝ V — a slower electron is wrongly given a shorter wavelength, and the square root is missing.
BThis takes λ ∝ √V: the dependence is inverted.
CCorrect: λ ∝ 1/√V and θ ∝ λ.
DThis takes λ ∝ 1/V, omitting the square root in p = √(2meV).
Syllabus understandingE.2 (HL) — diffraction of particles as evidence of the wave nature of matter; the de Broglie wavelength λ = h/p; the location of minimum intensity for diffracted particles based on their de Broglie wavelength; C.3 (HL) — single-slit diffraction θ = λ/bCommand term: Deduce
6E-1A-23
Stopping potential·E.2 Quantum physics (HL)
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDetermine
Light of different frequencies falls on a metal surface, and the stopping potential is measured for each frequency. The graph shows the results.
What is the work function of the metal?
Stopping potential against frequency of the incident light for one metal surface.Show mark scheme
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Notes
Step 1Einstein: eVs = hf − Φ, so Vs = (h/e)f − Φ/e: a straight line of gradient h/e.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2Extrapolating the line: it meets the frequency axis at f0 ≈ 5.5 × 1014 Hz (or the potential axis at ≈ −2.3 V).
ACorrect: the threshold frequency (or the intercept on the potential axis) gives Φ ≈ 2.3 eV.
BThis is the maximum kinetic energy of the photoelectrons at the highest frequency (eVs ≈ 2.7 eV), not the work function.
CThis is the photon energy hf at 12 × 1014 Hz, ignoring the kinetic energy of the electrons.
DThis adds instead of subtracts: hf + eVs = 5.0 + 2.7 eV (sign error in Einstein's equation).
Syllabus understandingE.2 (HL) — that photons of a certain frequency, known as the threshold frequency, are required to release photoelectrons; Einstein's explanation using the work function and the maximum kinetic energy of the photoelectrons as given by Emax = hf − Φ; Tool 3 — extrapolate graphs and interpret intercepts Command term: Determine
7E-1A-28
Intensity and the photoelectric current·E.2 Quantum physics (HL)
Paper 1AEasy1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
Monochromatic light falls on the cathode of the photocell shown. The pd of the supply is adjusted and the current is measured. The intensity of the light is then doubled, with the frequency unchanged.
Which row gives the effect on the maximum (saturation) current and on the stopping potential?
The supply can make the collector positive or negative with respect to the cathode.
Saturation currentStopping potential
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Notes
Step 1Doubling the intensity at fixed frequency doubles the number of photons per second, so twice as many electrons are emitted per second: the saturation current doubles.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2Each photon still has the same energy hf, so the maximum kinetic energy hf − Φ of the electrons is unchanged.
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Step 3The stopping potential Vs = (hf − Φ)/e therefore stays the same.
✓ 1
Answer A
Answer: A · 3 stages of work, one mark
Every option, and why
ACorrect: more photons, same photon energy.
BThe stopping potential depends only on the frequency (and the work function), not on the intensity.
CIntensity affects the number of electrons, not their energy; the current does change.
DThe current changes because the number of photons per second changes.
Syllabus understandingE.2 (HL) — the photoelectric effect as evidence of the particle nature of light; Einstein's explanation using the work function and the maximum kinetic energy of the photoelectrons as given by Emax = hf − Φ Command term: Deduce
8E-1A-35
The photoelectric effect·E.2 Quantum physics (HL)
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
In the classical wave model of light, the energy of the light is spread evenly over each wavefront and an electron in a metal surface absorbs energy continuously from the wave.
Two experiments are carried out on the same clean metal surface: Experiment 1: very dim light of a frequency well above the threshold frequency Experiment 2: very intense light of a frequency below the threshold frequency
Which row gives the predictions of the classical wave model?
Experiment 1Experiment 2
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Notes
Step 1Experiment 1: in the wave model a very dim beam delivers energy to each electron slowly, so an electron must absorb for some time before it has gained the work function: a delay is predicted.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2Experiment 2: in the wave model the energy delivered depends on the intensity, not the frequency, so a very intense beam soon gives electrons enough energy: emission is predicted at any frequency.
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Step 3Both predictions disagree with what is observed (immediate emission; no emission below the threshold frequency), which the photon model explains: one photon of energy hf gives all its energy to one electron.
✓ 1
Answer D
Answer: D · 3 stages of work, one mark
Every option, and why
AThese are the observations, which the photon model explains; the wave model predicts the opposite in both experiments.
BThis credits the wave model with a threshold frequency. In the wave model only the energy delivered per second matters, so intense light of any frequency should release electrons.
CThis credits the wave model with immediate emission. A dim wave spreads its energy over the whole surface, so each electron needs time to collect enough energy.
DCorrect: a dim wave needs time to deliver the work function, and an intense wave of any frequency eventually delivers it.
Syllabus understandingE.2 (HL) — the photoelectric effect as evidence of the particle nature of light; that photons of a certain frequency, known as the threshold frequency, are required to release photoelectrons from the metal; guidance: a discussion of which features of the photoelectric effect cannot be explained using the classical wave theory of light is required Command term: Deduce
9E-1A-36
The photoelectric effect·E.2 Quantum physics (HL)
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDetermine
Ultraviolet radiation of wavelength 250 nm falls on a metal surface and the maximum kinetic energy of the emitted electrons is 1.20 eV. The radiation is replaced by radiation of wavelength 350 nm. (h = 6.63 × 10−34 J s, c = 3.00 × 108 m s−1, e = 1.60 × 10−19 C)
What is the new maximum kinetic energy of the emitted electrons?
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Notes
Step 1hc/λ for 250 nm = 4.97 eV, so Φ = 4.97 − 1.20 = 3.77 eV.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2hc/λ for 350 nm = 3.55 eV.
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Step 33.55 eV < Φ: the frequency is below the threshold frequency, so no photoelectrons are released.
✓ 1
Answer A
Answer: A · 3 stages of work, one mark
Every option, and why
ACorrect: the 350 nm photons have less energy than the work function.
BThe maximum kinetic energy has been scaled in proportion to 1/λ (1.20 × 250/350), ignoring the work function.
CThe misconception that the maximum kinetic energy depends only on the metal (or on intensity), not on the frequency of the radiation.
DThe maximum kinetic energy has been scaled in proportion to λ (1.20 × 350/250) — inverted ratio.
Syllabus understandingE.2 (HL) — that photons of a certain frequency, known as the threshold frequency, are required to release photoelectrons from the metal; Emax = hf − ΦCommand term: Determine
10E-1A-37
Compton scattering·E.2 Quantum physics (HL)
Paper 1AEasy1 mark
Multiple choice · 1 mark3 steps to full marksIdentify
Photons are scattered by free electrons. Which graph shows how the shift Δλ in the wavelength of the scattered photons varies with the scattering angle θ?
Four graphs of the shift Δλ in the wavelength of the scattered photon against the scattering angle θ (λC is the Compton wavelength of the electron).Show mark scheme
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Notes
Step 1Δλ = (h/mec)(1 − cos θ) = λC(1 − cos θ).
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2θ = 0: Δλ = 0; θ = 90°: Δλ = λC; θ = 180°: Δλ = 2λC, and Δλ is never negative.
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Step 3Only graph D has these values.
✓ 1
Answer D
Answer: D · 3 stages of work, one mark
Every option, and why
AThis is λC cos θ: the "1 −" is missing, giving a shift at θ = 0 and a negative shift (shorter wavelength) for back-scattering.
BThis is λC sin θ: the trigonometric function has been swapped, putting the largest shift at 90°.
CThis is λC(1 + cos θ): a sign error that gives the largest shift for photons that are not deflected at all.
DCorrect: λC(1 − cos θ) rises from 0 to 2λC at 180°.
Syllabus understandingE.2 (HL) — that photons scatter off electrons with increased wavelength; the shift in photon wavelength after scattering off an electron as given by Δλ = (h/mec)(1 − cos θ) Command term: Identify
11E-1A-38
Matter waves·E.2 Quantum physics (HL)
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
An electron and a photon both have a wavelength of 0.20 nm.
Which row compares the momentum and the energy of the photon with the momentum and the kinetic energy of the electron?
MomentumEnergy
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Notes
Step 1For both, p = h/λ = 6.63 × 10−34/0.20 × 10−9 = 3.3 × 10−24 N s: the momenta are equal.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2Photon: E = pc = 9.9 × 10−16 J ≈ 6.2 keV.
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Step 3Electron: Ek = p²/2me = 6.0 × 10−18 J ≈ 38 eV, so the photon has much more energy.
✓ 1
Answer B
Answer: B · 3 stages of work, one mark
Every option, and why
AThe photon formula E = hc/λ has been applied to the electron as well.
BCorrect: equal momenta (p = h/λ); the photon energy pc is far larger than p²/2me.
CThe misconception that the photon, travelling at c, must have more momentum; de Broglie's relation gives the same momentum for the same wavelength.
DThe misconception that a particle with mass must carry more energy than a massless photon of the same wavelength.
Syllabus understandingE.2 (HL) — the de Broglie wavelength for particles as given by λ = h/p; that matter exhibits wave–particle duality Command term: Deduce
12E-1A-39
The photoelectric effect·E.2 Quantum physics (HL)
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDetermine
Monochromatic light of wavelength λ and power P falls on a photocathode. On average, only a fraction η of the incident photons each release one photoelectron, and every photoelectron released reaches the collector.
What is the photoelectric current?
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Step 1Photon energy = hc/λ, so the number of photons arriving per second is Pλ/hc.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2Electrons released per second = ηPλ/hc.
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Step 3Current = charge per second = ηPλe/hc.
✓ 1
Answer A
Answer: A · 3 stages of work, one mark
Every option, and why
ACorrect: I = e × (electrons per second) = ηPλe/hc.
BThis divides by the fraction η instead of multiplying, giving more electrons than photons.
CThis assumes every photon releases an electron: the fraction η has been ignored.
DThis writes the photon energy as hcλ instead of hc/λ.
Syllabus understandingE.2 (HL) — the photoelectric effect as evidence of the particle nature of light; E.1 — E = hf for a photon; B.5 — direct current as a flow of charge carriers I = Δq/ΔtCommand term: Determine
13E-1A-63
Intensity and the photoelectric current·E.2 Quantum physics (HL)
Paper 1AHard1 mark
Multiple choice · 1 mark4 steps to full marksDeduce
The same photocathode is illuminated in turn by two monochromatic light sources, P and Q. The graph shows how the photoelectric current I varies with the potential V of the anode relative to the cathode. For both sources, each incident photon has the same probability of releasing a photoelectron.
Which row compares the frequency and the intensity of light Q with those of light P?
Photoelectric current against anode potential for lights P and Q (graph drawn to scale)
Frequency of QIntensity of Q
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Step 1The current for Q falls to zero at −1.60 V and for P at −0.80 V: the stopping potential, and so Emax = eVs, is greater for Q.
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All 4 steps must be completed — there is no mark for a part-answer.
Step 2From Emax = hf − Φ with the same Φ, the photons of Q have the higher frequency.
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Step 3The saturation currents are equal, so equal numbers of photoelectrons, and therefore equal numbers of photons, arrive each second.
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Step 4Each photon of Q carries more energy, so the same photon rate means a greater energy per second per unit area: Q has the greater intensity.
✓ 1
Answer C
Answer: C · 4 stages of work, one mark
Every option, and why
AThis gets the frequency right but assumes that equal saturation currents mean equal intensities; the current fixes the photon rate, and each photon of Q carries more energy.
BThis reads the curve that reaches zero at the more negative potential as having the smaller stopping potential (lower frequency), and then reasons that a lower-frequency light needs a greater intensity to give the same current.
CCorrect: larger stopping potential → higher frequency; same photon rate with more energy per photon → greater intensity.
DThis misreads the stopping potentials and also equates equal currents with equal intensities.
Syllabus understandingE.2 (HL) — the photoelectric effect as evidence of the particle nature of light; Einstein's explanation using the work function and the maximum kinetic energy of the photoelectrons as given by Emax = hf − Φ; E.1 — E = hf for a photon Command term: Deduce
14E-1A-64
The photoelectric effect·E.2 Quantum physics (HL)
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksIdentify
Monochromatic light falls on a metal surface. The wavelength λ of the light is varied while the intensity is kept constant.
Which graph shows how the maximum kinetic energy Emax of the emitted electrons varies with λ?
Four graphs of the maximum kinetic energy of the photoelectrons against the wavelength of the incident light (λ₀ is the threshold wavelength).Show mark scheme
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Notes
Step 1Emax = hc/λ − Φ: this is not linear in λ; it falls steeply at short wavelengths and more gently near the threshold.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2Emax = 0 at the threshold wavelength λ₀ = hc/Φ.
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Step 3For λ > λ₀ the photons have too little energy and no electrons are emitted, so there is no value of Emax to plot (it is never negative). Only B shows this.
✓ 1
Answer B
Answer: B · 3 stages of work, one mark
Every option, and why
AThis treats Emax as a linear function of λ, as it is of frequency; Emax depends on 1/λ.
BCorrect: Emax = hc/λ − Φ, falling to zero at λ₀, with no emission at longer wavelengths.
CThis uses the right equation but extends it beyond λ₀, giving a negative kinetic energy; above λ₀ no electrons are emitted at all.
DThis copies the shape of the Emax–frequency graph onto a wavelength axis: longer wavelength means lower photon energy, not higher.
Syllabus understandingE.2 (HL) — that photons of a certain frequency, known as the threshold frequency, are required to release photoelectrons from the metal; Einstein's explanation using the work function and the maximum kinetic energy of the photoelectrons as given by Emax = hf − Φ Command term: Identify
15E-1A-85
Photoelectric effect and the wave model·E.2 Quantum physics (HL)
Paper 1AEasy1 mark
Multiple choice · 1 mark4 steps to full marksDeduce
Observations made in photoelectric experiments on clean metal surfaces include those listed below.
Which observation can be explained by the classical wave model of light?
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Notes
Step 1In the wave model a more intense wave delivers more energy per second to the surface, so more electrons can be released each second: A is consistent with the wave model (and with the photon model, in which more photons arrive each second).
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All 4 steps must be completed — there is no mark for a part-answer.
Step 2In the wave model a more intense wave also gives each electron more energy, so the maximum kinetic energy, and with it the stopping potential, should rise with intensity: B cannot be explained. In the photon model each photon still has energy hf, so eVs = hf − Φ is unchanged.
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Step 3In the wave model the energy delivered depends on the amplitude of the wave, not on its frequency, so there is no reason for Vs to depend on f, still less with the same gradient for every metal: C cannot be explained. In the photon model eVs = hf − Φ, so the gradient is h/e for every metal.
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Step 4In the wave model a dim wave delivers energy slowly, so an electron would need time to gather an energy Φ: a delay is predicted and D cannot be explained. In the photon model one photon gives all its energy hf to one electron at once.
✓ 1
Answer A
Answer: A · 4 stages of work, one mark
Every option, and why
ACorrect: a more intense wave delivers more energy per second, so the wave model also predicts that more electrons are released each second.
BThe wave model predicts that a more intense wave gives each electron more energy, and so a larger stopping potential; an unchanged Vs needs photons of fixed energy hf.
CIn the wave model the energy given to an electron depends on the amplitude of the wave, not on its frequency; the common gradient h/e of the Vs–f line is evidence for E = hf.
DA dim wave would need time to give one electron an energy Φ, so the wave model predicts a delay; immediate emission is explained by one photon giving all its energy to one electron.
Syllabus understandingE.2 — the photoelectric effect as evidence of the particle nature of light; Einstein’s explanation using the work function and the maximum kinetic energy of the photoelectrons as given by Emax = hf − Φ where Φ is the work function of the metal; Guidance: a discussion of which features of the photoelectric effect cannot be explained using the classical wave theory of light is required Command term: Deduce
16E-1A-86
Compton scattering·E.2 Quantum physics (HL)
Paper 1AEasy1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
An X-ray photon travelling in a vacuum is scattered by a free electron that is initially at rest.
Which row compares the scattered photon with the incident photon?
SpeedFrequency
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Notes
Step 1The electron recoils, so it gains kinetic energy; by conservation of energy the scattered photon has less energy than the incident photon.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2E = hf, so the frequency of the scattered photon is lower and its wavelength is longer: Δλ = (h/mec)(1 − cos θ) ≥ 0.
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Step 3Every photon travels at c in a vacuum: the photon loses energy through a lower frequency, not a lower speed.
✓ 1
Answer D
Answer: D · 3 stages of work, one mark
Every option, and why
AThis treats the photon like a ball that slows down when it loses energy. A photon always travels at c in a vacuum; its energy is fixed by its frequency.
BThe frequency is right, but the speed of a photon in a vacuum does not change.
CThis has the energy transfer the wrong way round: the electron starts at rest and gains kinetic energy, so the photon cannot gain energy.
DCorrect: the photon keeps the speed c and gives up energy to the electron, so its frequency is lower.
Syllabus understandingE.2 — Compton scattering of light by electrons as additional evidence of the particle nature of light; that photons scatter off electrons with increased wavelength; E.1 — the energy of a photon E = hfCommand term: Deduce
17E-1A-87
De Broglie wavelength of accelerated particles·E.2 Quantum physics (HL)
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
A proton and an alpha particle are each accelerated from rest through the same potential difference. The alpha particle has 4 times the mass and 2 times the charge of the proton.
What is (de Broglie wavelength of the proton)/(de Broglie wavelength of the alpha particle)?
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Notes
Step 1The kinetic energy gained is qV, so Ek ∝ q.
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All 3 steps must be completed — there is no mark for a part-answer.
AThis includes the doubled charge (twice the kinetic energy) but ignores the mass in p = √(2mEk): √2.
BThis assumes that both particles gain the same kinetic energy, ignoring the doubled charge: √4 = 2.
CCorrect: λ ∝ 1/√(mq), and the alpha particle has 8 times the product mq.
DThis omits the square root in p = √(2mEk): 4 × 2 = 8.
Syllabus understandingE.2 — the de Broglie wavelength for particles as given by λ = h/p; D.2 (HL) — the work done moving a charge, W = qΔVe; A.3 — Ek = p²/2mCommand term: Deduce
18E-1A-88
Compton scattering·E.2 Quantum physics (HL)
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDetermine
X-rays of a single wavelength, 56.0 pm, are scattered by a block of graphite. A detector placed at an angle θ to the incident beam measures the intensity of the scattered X-rays at different wavelengths. The graph shows the results. The peak at 56.0 pm is due to X-rays scattered by electrons that are tightly bound in atoms; the other peak is due to scattering by electrons that behave as free electrons.
h/mec = 2.43 pm. What is θ?
Intensity of the X-rays scattered through angle θ against wavelength (drawn to scale).Show mark scheme
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Notes
Step 1The second peak is at 57.6 pm, so Δλ = 57.6 − 56.0 = 1.6 pm.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2Δλ = (h/mec)(1 − cos θ), so 1 − cos θ = 1.6/2.43 = 0.66.
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Step 3cos θ = 0.34, so θ = 70°.
✓ 1
Answer C
Answer: C · 3 stages of work, one mark
Every option, and why
AThis uses (1 − sin θ) in place of (1 − cos θ): sin θ = 0.34 gives 20°.
BThis sets cos θ = Δλ/(h/mec) = 0.66, leaving out the “1 −” of the formula: 49°.
CCorrect: Δλ = 1.6 pm, so 1 − cos θ = 0.66 and θ = 70°.
DThis uses (1 + cos θ): cos θ = −0.34 gives 110°.
Syllabus understandingE.2 — Compton scattering of light by electrons as additional evidence of the particle nature of light; that photons scatter off electrons with increased wavelength; the shift in photon wavelength after scattering off an electron as given by λf − λi = Δλ = (h/mec)(1 − cos θ) Command term: Determine
19E-1A-89
Threshold wavelength·E.2 Quantum physics (HL)
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDetermine
The threshold wavelength for a metal surface is λ0. Light of wavelength λ, where λ < λ0, falls on the surface. The mass of an electron is me.
What is the maximum speed of the emitted photoelectrons?
All 3 steps must be completed — there is no mark for a part-answer.
Step 2½mev² = Emax.
—
Step 3v = √(2hc(λ0 − λ)/(meλλ0)).
✓ 1
Answer D
Answer: D · 3 stages of work, one mark
Every option, and why
AThis omits the ½ in the kinetic energy, writing Emax = mev².
BThis replaces 1/λ − 1/λ0 by 1/(λ0 − λ), which is not algebraically equal.
CThis subtracts the photon energy from the work function (hc/λ0 − hc/λ), a sign error that makes the expression under the root negative when λ < λ0.
DCorrect: Emax = hc(1/λ − 1/λ0) = ½mev².
Syllabus understandingE.2 — that photons of a certain frequency, known as the threshold frequency, are required to release photoelectrons from the metal; Einstein’s explanation using the work function and the maximum kinetic energy of the photoelectrons as given by Emax = hf − Φ where Φ is the work function of the metal; A.3 — Ek = ½mv² Command term: Determine
20E-1A-90
Wave–particle duality·E.2 Quantum physics (HL)
Paper 1AEasy1 mark
Multiple choice · 1 mark3 steps to full marksExplain
A pollen grain of mass 2.0 × 10−14 kg drifts at a speed of 1.0 mm s−1 towards a gap of width 0.50 mm in a filter.
Why is no diffraction of the pollen grain observed as it passes through the gap?
Show mark scheme
Marking point
Mark
Notes
Step 1p = mv = 2.0 × 10−14 × 1.0 × 10−3 = 2.0 × 10−17 kg m s−1.
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 3Diffraction is noticeable only when the wavelength is comparable with the gap; 3 × 10−17 m is about 1013 times smaller than 0.50 mm, so the spreading is far too small to detect.
✓ 1
Answer B
Answer: B · 3 stages of work, one mark
Every option, and why
AAll matter has a de Broglie wavelength λ = h/p; wave–particle duality is not limited to elementary particles (molecules of many atoms have been diffracted).
BCorrect: the wavelength is about 1013 times smaller than the gap, so the diffraction is negligible.
CThis inverts de Broglie’s relation, calculating p/h = 3 × 1016 m−1 and treating it as a wavelength.
DA slower particle has less momentum and so a longer wavelength; diffraction does not need a high speed (slow neutrons and atoms are diffracted).
Syllabus understandingE.2 — that matter exhibits wave–particle duality; the de Broglie wavelength for particles as given by λ = h/p; diffraction of particles as evidence of the wave nature of matter; C.3 — diffraction through an aperture Command term: Explain
21E-1A-91
Threshold frequency·E.2 Quantum physics (HL)
Paper 1AEasy1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
Light of wavelength 650 nm does not release any photoelectrons from a metal surface, however great its intensity. Light of wavelength 450 nm does release photoelectrons from the same surface.
All 3 steps must be completed — there is no mark for a part-answer.
Step 2450 nm: 4.42 × 10−19 J = 2.76 eV.
—
Step 3No emission at 650 nm means Φ > 1.91 eV; emission at 450 nm means Φ < 2.76 eV. Only 2.2 eV lies in this range.
✓ 1
Answer B
Answer: B · 3 stages of work, one mark
Every option, and why
A1.6 eV is less than the energy of a 650 nm photon (1.91 eV), so 650 nm light would also release electrons: this value fits only the 450 nm observation.
BCorrect: the work function lies between the photon energies 1.91 eV and 2.76 eV.
C3.1 eV is more than the energy of a 450 nm photon (2.76 eV), so 450 nm light would release no electrons: this value fits only the 650 nm observation.
DThis adds the two photon energies (1.91 + 2.76 eV = 4.7 eV); one photon gives its energy to one electron, and photons of different wavelengths do not combine.
Syllabus understandingE.2 — that photons of a certain frequency, known as the threshold frequency, are required to release photoelectrons from the metal; Einstein’s explanation using the work function and the maximum kinetic energy of the photoelectrons as given by Emax = hf − Φ where Φ is the work function of the metal; E.1 — E = hfCommand term: Deduce
22E-1A-92
Compton scattering and momentum·E.2 Quantum physics (HL)
Paper 1AEasy1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
A photon travelling in the direction shown is scattered by a free electron at rest at O. The scattered photon travels at 60° to the original direction. Arrows A, B, C and D are drawn from O.
Which arrow could show the direction of motion of the recoiling electron?
Scattering of a photon by a free electron initially at rest at O (not to scale).Show mark scheme
Marking point
Mark
Notes
Step 1Momentum is conserved. The electron starts at rest, so its momentum after the scattering equals (momentum of incident photon) − (momentum of scattered photon).
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 2Along the original direction the scattered photon carries only (h/λf) cos 60° of the incident momentum h/λi, and λf > λi, so the electron must move forward.
—
Step 3At right angles to it the scattered photon has a momentum component upwards, so the electron must have an equal component downwards. Only D points forward and below the original line.
✓ 1
Answer D
Answer: D · 3 stages of work, one mark
Every option, and why
AThis conserves momentum only along the original direction; the upward momentum of the scattered photon must be balanced by a downward component of the electron’s momentum.
BThis puts the electron on the same side as the scattered photon, so the perpendicular components of momentum would add instead of cancelling.
CThis balances only the momentum of the scattered photon (the electron moving exactly opposite to it) and forgets the momentum that the incident photon brought along the original direction.
DCorrect: the electron must move forward and below the original direction; for a 100 keV photon, for example, the angle is about 55°.
Syllabus understandingE.2 — Compton scattering of light by electrons as additional evidence of the particle nature of light; that photons scatter off electrons with increased wavelength; A.2 — conservation of linear momentum, collisions in two dimensions (HL) Command term: Deduce
23E-1A-93
Compton scattering and the wave model·E.2 Quantum physics (HL)
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
X-rays of a single wavelength are scattered by electrons in a block of carbon, and a detector measures the wavelength of the X-rays scattered through 90°. In the classical wave model, the oscillating electric field of the X-rays makes each electron oscillate at the frequency of the wave, and each oscillating electron emits radiation at the frequency of its own oscillation.
Which row gives the prediction of the classical wave model and the observed result for the X-rays scattered by free electrons?
Prediction of the classical wave modelObserved result
Show mark scheme
Marking point
Mark
Notes
Step 1Classical model: each electron is forced to oscillate at the frequency of the incident wave and re-emits at that frequency, so the scattered wavelength equals the incident wavelength at every angle.
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 2Observed: the X-rays scattered by free electrons have a longer wavelength, shifted by Δλ = (h/mec)(1 − cos 90°) = 2.43 pm.
—
Step 3The shift is explained by a photon colliding with an electron and giving it kinetic energy and momentum, so Compton scattering is evidence for the particle nature of light.
✓ 1
Answer A
Answer: A · 3 stages of work, one mark
Every option, and why
ACorrect: the wave model predicts no change of wavelength, but a longer wavelength is observed.
BThis gives the wave model the energy loss of the photon model. A forced oscillator re-radiates at the driving frequency, so the classical model predicts no change of wavelength.
CThe classical prediction is right, but the X-rays scattered by free electrons are observed at a longer wavelength: this is the Compton shift.
DThis reverses the two: the shift is observed, but it is not predicted by the wave model.
Syllabus understandingE.2 — Compton scattering of light by electrons as additional evidence of the particle nature of light; that photons scatter off electrons with increased wavelength; the shift in photon wavelength after scattering off an electron as given by λf − λi = Δλ = (h/mec)(1 − cos θ) Command term: Deduce
24E-1A-94
Stopping potential·E.2 Quantum physics (HL)
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
When light of wavelength λ1 falls on a metal surface, the stopping potential is V1. With light of a longer wavelength λ2, the stopping potential is V2.
Which expression gives the Planck constant?
Show mark scheme
Marking point
Mark
Notes
Step 1For each wavelength eVs = Emax = hc/λ − Φ.
—
All 3 steps must be completed — there is no mark for a part-answer.
AThis replaces 1/λ1 − 1/λ2 by 1/(λ2 − λ1), which is not algebraically equal.
BThis divides by e instead of multiplying: the kinetic energy of an electron stopped by Vs is eVs, not Vs/e.
CCorrect: eliminating the work function between the two equations gives this expression.
DThis omits c, writing the photon energy as h/λ instead of hc/λ.
Syllabus understandingE.2 — the photoelectric effect as evidence of the particle nature of light; Einstein’s explanation using the work function and the maximum kinetic energy of the photoelectrons as given by Emax = hf − Φ where Φ is the work function of the metal Command term: Deduce
25E-1A-95
Diffraction of electrons·E.2 Quantum physics (HL)
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDetermine
A beam of electrons, accelerated from rest through a potential difference V, is directed at a single slit of width 80 nm. A detector counts the electrons arriving each minute at different angles θ from the central direction. The graph shows the results.
What is V?
Electrons detected per minute against angle θ from the central direction (drawn to scale).Show mark scheme
Marking point
Mark
Notes
Step 1The first minima are at θ = ±3.0 mrad. For a single slit θ = λ/b, so λ = 80 × 10−9 × 3.0 × 10−3 = 2.4 × 10−10 m.
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 2p = h/λ = 2.76 × 10−24 kg m s−1; eV = p²/2me = 4.19 × 10−18 J.
—
Step 3V = 26 V.
✓ 1
Answer B
Answer: B · 3 stages of work, one mark
Every option, and why
AThis takes the angle between the two first minima, 6.0 mrad, as θ: λ is doubled, so V is a quarter of the true value.
BCorrect: λ = bθ = 2.4 × 10−10 m and V = p²/(2mee) = 26 V.
CThis omits the ½ in Ek = p²/2me, writing eV = p²/me: 52 V.
DThis treats the electron as a photon of the same wavelength, eV = hc/λ: 5.2 kV.
Syllabus understandingE.2 — diffraction of particles as evidence of the wave nature of matter; the de Broglie wavelength for particles as given by λ = h/p; Guidance: a description of a scattering experiment including the location of minimum intensity for the diffracted particles based on their de Broglie wavelength is required; C.3 (HL) — single-slit diffraction, θ = λ/b; D.2 (HL) — W = qΔVeCommand term: Determine
26E-1A-96
Compton scattering·E.2 Quantum physics (HL)
Paper 1AHard1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
Photon X is scattered through 60° by a free electron at rest, and then through a further 60° by a second free electron at rest, so that it finally travels at 120° to its original direction. Photon Y, with the same initial wavelength as X, is scattered through 120° by a single free electron at rest.
What is (total increase in the wavelength of X)/(increase in the wavelength of Y)?
Show mark scheme
Marking point
Mark
Notes
Step 1Δλ = (h/mec)(1 − cos θ) does not depend on the wavelength of the photon, so each 60° scattering of X gives the same shift.
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 2X: 2 × (1 − cos 60°) = 2 × 0.5 = 1.0 (in units of h/mec). Y: 1 − cos 120° = 1.5.
—
Step 3Ratio = 1.0/1.5 = 2/3.
✓ 1
Answer A
Answer: A · 3 stages of work, one mark
Every option, and why
ACorrect: two small-angle scatterings give a smaller total shift than one scattering through the same total angle, because 1 − cos θ rises more and more steeply as θ increases.
BThis assumes that the shift depends only on the final direction of the photon, or is proportional to the total angle turned (60° + 60° = 120°); the shift varies as 1 − cos θ, which is not proportional to θ.
CThis inverts the ratio, giving (increase for Y)/(increase for X).
DThis uses sin θ in place of 1 − cos θ: 2 sin 60°/sin 120° = 2.
Syllabus understandingE.2 — that photons scatter off electrons with increased wavelength; the shift in photon wavelength after scattering off an electron as given by λf − λi = Δλ = (h/mec)(1 − cos θ) Command term: Deduce
27E-1A-97
Stopping potential·E.2 Quantum physics (HL)
Paper 1AHard1 mark
Multiple choice · 1 mark3 steps to full marksDetermine
A lamp emits light with a continuous range of wavelengths from 250 nm to 700 nm. The light passes through the glass wall of a vacuum photocell, which absorbs all wavelengths shorter than 350 nm and transmits all longer wavelengths. The light then falls on a cathode of work function 2.25 eV.
What is the smallest potential difference between the anode and the cathode that stops the photoelectric current?
Show mark scheme
Marking point
Mark
Notes
Step 1The current stops only when the fastest photoelectrons are turned back, so the stopping potential is set by the photons of greatest energy, i.e. the shortest wavelength, that reach the cathode.
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 2The glass absorbs everything below 350 nm, so the shortest wavelength reaching the cathode is 350 nm: hc/λ = 5.68 × 10−19 J = 3.55 eV.
—
Step 3Emax = 3.55 − 2.25 = 1.30 eV, so Vs = 1.3 V.
✓ 1
Answer A
Answer: A · 3 stages of work, one mark
Every option, and why
ACorrect: the shortest wavelength that reaches the cathode is 350 nm, giving Emax = 3.55 − 2.25 = 1.3 eV.
BThis uses the shortest wavelength emitted by the lamp, 250 nm (4.97 eV), ignoring the absorption by the glass.
CThis is the energy of a 350 nm photon in eV, without subtracting the work function.
DThis adds the work function to the photon energy (3.55 + 2.25 eV), a sign error in Emax = hf − Φ.
Syllabus understandingE.2 — that photons of a certain frequency, known as the threshold frequency, are required to release photoelectrons from the metal; Einstein’s explanation using the work function and the maximum kinetic energy of the photoelectrons as given by Emax = hf − Φ where Φ is the work function of the metal; E.1 — E = hfCommand term: Determine
28E-1B-02
Stopping potential·E.2 Quantum physics (HL)
Paper 1BMedium6 marks
Data-based question5 steps to full marksDetermine
In a vacuum photocell, light from a white source passes through one of six narrow-band filters and falls on a cathode coated with potassium. For each filter the student increases the reverse potential difference between anode and cathode until a sensitive ammeter in the circuit just reads zero; this is the stopping potential Vs. The frequency f passed by each filter is given by the manufacturer. Because the current falls to zero gradually, each Vs is uncertain by ±0.10 V.
The graph shows the data with error bars and the line of best fit. (e = 1.60 × 10−19 C)
f / 1014 Hz
6.0
7.0
8.0
9.0
10.0
11.0
Vs / V
0.26
0.62
1.08
1.51
1.87
2.31
Stopping potential against frequency (graph drawn to scale)
(a)
(i)
Determine, using the line of best fit, a value for the Planck constant.
(2)
(b)
(i)
Draw the lines of maximum and minimum gradient that are consistent with the error bars. Hence determine the absolute uncertainty in your value of the Planck constant.
(2)
(c)
(i)
Determine, by extrapolating the line of best fit, the work function of potassium in eV.
(1)
(d)
(i)
The accepted value of the Planck constant is 6.63 × 10−34 J s. Comment on your result.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
Gradient from two well-separated points on the line = 4.12 × 10−15 V s
✓ 1
Accept 4.0–4.2 × 10−15 V s.
h = e × gradient = 1.60 × 10−19 × 4.12 × 10−15 = 6.6 × 10−34 J s
✓ 1
Allow ECF from the gradient. Accept 6.4–6.7 × 10−34 J s.
Part (b)(i)
Both lines pass through every error bar; gradients ≈ 4.5 × 10−15 V s and 3.7 × 10−15 V s
✓ 1
Accept 4.4–4.6 × 10−15 and 3.6–3.9 × 10−15 V s.
Δh = e(gmax − gmin)/2 = ±0.6 × 10−34 J s
✓ 1
Allow ECF from (a) if the fractional route Δh/h = Δg/g is used. Accept ±0.5 to ±0.7 × 10−34 J s; must be given to 1 s.f.
Part (c)(i)
Intercept on the Vs axis = −2.2 V, so Φ = 2.2 eV
✓ 1
Accept 2.1–2.4 eV. ALT: f0 = 5.4 × 1014 Hz from the f axis and Φ = hf0/e using the value from (a); Allow ECF from (a).
Part (d)(i)
6.63 × 10−34 J s lies inside 6.6 ± 0.6 (× 10−34 J s), so the result is consistent with the accepted value
✓ 1
Allow ECF from (a) and (b). The conclusion must agree with the candidate's own value and uncertainty.
Answers: (a)(i) 6.6 × 10−34 J s · (b)(i) ±0.6 × 10−34 J s · (c)(i) 2.2 eV (the remaining parts are explanations — see the table above)
Syllabus understandingE.2 (HL) — the photoelectric effect as evidence of the particle nature of light; the photoelectric equation Emax = hf − Φ; the threshold frequency; Tools 3 — error bars, lines of maximum and minimum gradient, uncertainty in a gradient, extrapolation to an intercept Command term: Determine
29E-1B-06
Electron diffraction·E.2 Quantum physics (HL)
Paper 1BMedium6 marks
Data-based question5 steps to full marksDetermine
In an electron diffraction tube, electrons are accelerated from rest through a potential difference V and then pass through a very thin layer of polycrystalline graphite. Rings appear on a screen a distance L = 0.135 m beyond it. For each V the student measures the radius r of the innermost ring. For small angles r = Lλ/d, where λ is the de Broglie wavelength of the electrons and d is the spacing of the planes of atoms that produce the ring.
The graph shows r against V−1/2 with a line of best fit. (h = 6.63 × 10−34 J s, me = 9.11 × 10−31 kg, e = 1.60 × 10−19 C)
V / V
2000
3000
4000
5000
6000
r / cm
1.75
1.41
1.24
1.09
1.01
V−1/2 / 10−3 V−1/2
22.36
18.26
15.81
14.14
12.91
Ring radius against V−1/2 (graph drawn to scale)
(a)
(i)
Determine the gradient of the graph, giving its unit.
(2)
(b)
(i)
Determine d.
(3)
(c)
(i)
Calculate the de Broglie wavelength of the electrons when V = 4000 V and use it to comment on whether the small-angle approximation is justified.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
Gradient = 0.78 (from e.g. 1.95 cm at 25 × 10−3 V−1/2)
✓ 1
Accept 0.76–0.80 m V1/2 (76–80 cm V1/2).
Unit: m V1/2 (or cm V1/2 with a consistent value)
✓ 1
Part (b)(i)
eV = p²/2me, so p = √(2meeV) and λ = h/√(2meeV)
✓ 1
Work done by the field equals the kinetic energy gained.
Hence r = [Lh/(d√(2mee))] V−1/2: gradient = Lh/(d√(2mee))
λ = 1.9 × 10−11 m, so λ/d = 0.091 rad (≈ 5°): small, so the approximation is justified
✓ 1
Allow ECF from (b). Both the wavelength and a comparison of λ/d (or the angle) with 1 rad are needed.
Answers: (a)(i) 0.78 m V1/2 · (b)(i) 2.1 × 10−10 m · (c)(i) 1.9 × 10−11 m (the remaining parts are explanations — see the table above)
Syllabus understandingE.2 (HL) — the de Broglie wavelength λ = h/p; diffraction of particles as evidence of the wave nature of matter; D.2 — the work done on a charge moved through a potential difference, W = qV; C.3 — diffraction by a regular array; Tools 3 — gradient of a linearised graph with its unit Command term: Determine
30E-1B-13
Compton scattering·E.2 Quantum physics (HL)
Paper 1BEasy6 marks
Data-based question4 steps to full marksDetermine
Monochromatic X-rays of wavelength 70.9 pm are directed at a block of graphite. A crystal spectrometer on a rotating arm measures the wavelength λf of the X-rays scattered through angle θ; each value is uncertain by ±0.15 pm. The Compton formula predicts that λf = λi + (h/mec)(1 − cos θ), where λi is the incident wavelength.
The graph shows λf against (1 − cos θ) with error bars and the line of best fit. (h = 6.63 × 10−34 J s, c = 3.00 × 108 m s−1)
θ / °
1 − cos θ
λf / pm
30
0.134
71.3
60
0.500
72.1
90
1.000
73.4
120
1.500
74.5
150
1.866
75.5
180
2.000
75.7
Scattered wavelength against (1 − cos θ) (graph drawn to scale)
(a)
(i)
Determine the gradient of the line.
(2)
(b)
(i)
Determine the mass of the electron from your answer to (a), and compare it with the accepted value 9.11 × 10−31 kg.
(2)
(c)
(i)
Explain why the reading at θ = 30° contributes least to the determination of the gradient, and suggest how the student could improve the experiment using the same spectrometer.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
Two well-separated points on the line, e.g. (0, 71.0) and (2.00, 75.7)
✓ 1
Gradient = 2.40 pm
✓ 1
Accept 2.30–2.50 pm (2.4 × 10−12 m).
Part (b)(i)
me = h/(gradient × c) = 6.63 × 10−34/(2.40 × 10−12 × 3.00 × 108) = 9.2 × 10−31 kg
✓ 1
Allow ECF from (a). Accept 8.8–9.6 × 10−31 kg.
About 1 % above the accepted value — close agreement, supporting the Compton model
✓ 1
The comparison must use the candidate's own value.
Part (c)(i)
Shift at 30° = gradient × 0.134 = 0.32 pm, so ±0.15 pm is about 47 % of the shift (only ≈ 3 % at 180°)
✓ 1
Allow ECF from (a).
Take more (repeated) readings at large angles, 120°–180°, where the shift is largest, and fewer at small angles
✓ 1
Do not accept "use a more precise spectrometer".
Answers: (a)(i) 2.4 pm · (b)(i) 9.2 × 10−31 kg (the remaining parts are explanations — see the table above)
Syllabus understandingE.2 (HL) — Compton scattering of light by electrons as additional evidence of the particle nature of light; λf − λi = Δλ = (h/mec)(1 − cos θ); Tools 3 — gradient and intercept, comparison with an accepted value; Inquiry 2 — improving precision with the same instrument Command term: Determine
31E-1B-14
Matter waves·E.2 Quantum physics (HL)
Paper 1BHard6 marks
Data-based question4 steps to full marksDetermine
A beam of fullerene molecules, known to be either C60 (mass 720 u) or C70 (mass 840 u), passes through a velocity selector and then through a single slit of width b = 0.20 μm. A detector scanned across the beam L = 1.50 m beyond the slit records a diffraction pattern; the distance y from its centre to the first minimum is measured for different molecular speeds v. The first minimum is at angle θ ≈ λ/b, so that y = (hL/mb)(1/v) for molecules of mass m.
The graph shows y against 1/v with the line of best fit. (h = 6.63 × 10−34 J s, 1 u = 1.66 × 10−27 kg)
v / m s−1
1/v / 10−3 s m−1
y / μm
100
10.00
43.5
125
8.00
34.5
150
6.67
29.5
175
5.71
25.1
200
5.00
22.8
225
4.44
19.9
250
4.00
18.5
y against 1/v (graph drawn to scale)
(a)
(i)
Determine the gradient of the line in SI units.
(2)
(b)
(i)
Determine the mass of one molecule in u and deduce which fullerene was used.
(2)
(c)
(i)
Another student uses only the reading at 250 m s−1 and assumes that y is proportional to 1/v. Determine the mass he obtains, and explain why the method in (b) is more reliable.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
Two well-separated points on the line, e.g. (0, 1.7 μm) and (10.0 × 10−3, 43.2 μm)
Gradient = 18.5/4.00 = 4.62 μm per 10−3 s m−1, so m = 648 u — matching neither fullerene
✓ 1
Accept 640–660 u.
The line has an intercept of ≈ 1.7 μm (a zero offset, e.g. the centre of the pattern located wrongly); a single point through the origin includes this offset, whereas the gradient of the full line in (b) is unaffected by it
✓ 1
Allow ECF from (b). Accept intercept 1.2–2.0 μm.
Answers: (a)(i) 4.2 × 10−3 m² s−1 · (b)(i) 721 u, C60 · (c)(i) 648 u (the remaining parts are explanations — see the table above)
Syllabus understandingE.2 (HL) — diffraction of particles as evidence of the wave nature of matter, including the location of minimum intensity for the diffracted particles based on their de Broglie wavelength; λ = h/p; C.3 — single-slit diffraction, θ = λ/b; Tools 3 — gradient in SI units, intercept as a systematic error, using the full data set Command term: Determine
32E-1B-27
Threshold wavelength from filters·E.2 Quantum physics (HL)
Paper 1BEasy6 marks
Data-based question6 steps to full marksDetermine
A vacuum photocell has a cathode coated with an unknown metal. Light from a lamp passes through one of eight interference filters, each transmitting a narrow band of wavelengths centred on λ, and falls on the cathode. The anode is kept positive relative to the cathode and a picoammeter measures the current I. Each reading is uncertain by ±0.1 pA. With the lamp switched off, the picoammeter reads 0.6 pA because of a small leakage current through the insulation of the photocell. The table shows the results with the lamp switched on.
(h = 6.63 × 10−34 J s, c = 3.00 × 108 m s−1, e = 1.60 × 10−19 C)
λ / nm
400
425
450
475
500
525
550
575
I / pA
55.8
30.0
13.9
4.8
1.0
0.6
0.6
0.6
(a)
(i)
A student concludes that some photoelectrons are emitted at every wavelength in the table. Explain why this conclusion is incorrect.
(1)
(ii)
Deduce the range of wavelengths within which the threshold wavelength lies.
(1)
(b)
(i)
Determine, in eV, the work function of the metal and its absolute uncertainty.
(2)
(ii)
The supplier states that the work function of the coating is 2.30 eV. Comment on this statement.
(1)
(c)
(i)
Suggest how the student could reduce the uncertainty in the work function, using the same photocell, lamp and picoammeter.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
For λ ≥ 525 nm the reading is 0.6 pA, the same as with the lamp switched off (within ±0.1 pA): this is the leakage current, so no photoelectrons are emitted at these wavelengths
✓ 1
A comparison with the 0.6 pA reading with the lamp off is needed.
Part (a)(ii)
Between 500 nm and 525 nm: at 500 nm the reading, 1.0 pA, exceeds 0.6 pA by more than the combined uncertainty of ±0.2 pA
✓ 1
Allow ECF from (a)(i).
Part (b)(i)
hc/λ for 525 nm = 2.37 eV and for 500 nm = 2.49 eV
✓ 1
Allow ECF from (a)(ii).
Φ = 2.43 ± 0.06 eV
✓ 1
Mid-value and half the range. Accept ±0.06 or ±0.07 eV; the uncertainty must be given to 1 s.f.
Part (b)(ii)
Not consistent: 2.30 eV lies outside the range 2.37–2.49 eV (a work function of 2.30 eV would give a threshold of 540 nm, so emission would be seen at 525 nm)
✓ 1
Allow ECF from (b)(i): the conclusion must agree with the candidate’s own range.
Part (c)(i)
Use further filters with centre wavelengths between 500 nm and 525 nm (for example every 5 nm), so that the threshold wavelength is bracketed more closely
✓ 1
“Repeat the readings” or “use a more sensitive meter”: [0].
Answers: (a)(ii) 500 nm to 525 nm · (b)(i) 2.43 ± 0.06 eV · (b)(ii) not consistent (the remaining parts are explanations — see the table above)
Syllabus understandingE.2 — that photons of a certain frequency, known as the threshold frequency, are required to release photoelectrons from the metal; Einstein’s explanation using the work function and the maximum kinetic energy of the photoelectrons as given by Emax = hf − Φ where Φ is the work function of the metal; Tools 1 and 3 — a systematic (zero) offset, the absolute uncertainty of a value found from bracketing readings; Inquiry 3 — improving an investigation with the same apparatus Command term: Determine
33E-1B-28
Photoelectric current and distance·E.2 Quantum physics (HL)
Paper 1BMedium7 marks
Data-based question5 steps to full marksDetermine
A small lamp fitted with a filter emits light of wavelength 436 nm. The light falls on the cathode of a vacuum photocell, and the anode is kept at a positive potential large enough for every emitted photoelectron to reach it. The photocurrent I is measured for different distances d, measured with a metre rule from the front of the lamp housing to the cathode. The lamp filament, which acts as a point source, is a fixed but unknown distance x0 behind the front of the housing, so the filament is a distance d + x0 from the cathode.
The table shows the results. The graph shows I−1/2 against d with the line of best fit. (h = 6.63 × 10−34 J s, c = 3.00 × 108 m s−1, e = 1.60 × 10−19 C)
d / cm
I / nA
I−1/2 / nA−1/2
10
79.7
0.112
15
40.5
0.157
20
24.8
0.201
25
16.4
0.247
30
11.8
0.291
40
6.9
0.381
I−1/2 against d with the line of best fit (drawn to scale).
(a)
(i)
The number of photons reaching the cathode each second is inversely proportional to (d + x0)². Explain why a graph of I−1/2 against d is expected to be a straight line.
(2)
(ii)
Determine x0.
(2)
(b)
(i)
At d = 20 cm the power of the light reaching the cathode is 0.32 μW. Determine the fraction of the photons reaching the cathode that release a photoelectron.
(2)
(c)
(i)
The lamp is moved from d = 10 cm to d = 40 cm. State and explain the effect on the stopping potential of the photocell.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
Each photon can release at most one electron, with a fixed probability, so I is proportional to the number of photons arriving per second: I = k/(d + x0)²
✓ 1
Reference to one photon releasing one electron is needed.
So I−1/2 = (d + x0)/√k, a linear function of d (gradient 1/√k, meeting the d-axis at −x0)
✓ 1
Part (a)(ii)
The line of best fit is extrapolated to I−1/2 = 0, or gradient = 0.0090 nA−1/2 cm−1 and vertical intercept = 0.022 nA−1/2
✓ 1
Accept a gradient of 0.0086–0.0094 nA−1/2 cm−1 and an intercept of 0.018–0.026 nA−1/2.
Electrons per second = 24.8 × 10−9/1.60 × 10−19 = 1.55 × 1011 s−1; fraction = 0.22
✓ 1
Accept 0.21–0.23.
Part (c)(i)
No change: each photon still has energy hc/λ (436 nm), so Emax = hf − Φ and the stopping potential are unchanged; only the number of photons per second decreases
✓ 1
Both the conclusion and the reason are needed.
Answers: (a)(ii) 2.5 cm · (b)(i) 0.22 · (c)(i) no change (the remaining parts are explanations — see the table above)
Syllabus understandingE.2 — the photoelectric effect as evidence of the particle nature of light; Einstein’s explanation using the work function and the maximum kinetic energy of the photoelectrons as given by Emax = hf − Φ where Φ is the work function of the metal; B.1 — apparent brightness b = L/4πd² (inverse-square law); Tools 3 — linearising a relationship, extrapolation to an intercept, an intercept as a systematic error Command term: Determine
34E-1B-29
Photoelectrons in a magnetic field·E.2 Quantum physics (HL)
Paper 1BHard7 marks
Data-based question5 steps to full marksDetermine
Monochromatic ultraviolet light of wavelength λ falls on a small spot on a flat cathode in an evacuated chamber. A uniform magnetic field of flux density B = 1.00 × 10−4 T acts parallel to the cathode surface. Photoelectrons leaving the spot move in circular arcs in the field and return to the plane of the cathode, which is covered with a fluorescent coating around the spot. Measured in the direction at right angles to the field, the glow extends furthest from the spot for the fastest photoelectrons emitted at right angles to the surface; this greatest distance D is equal to the diameter of their circular path. D is measured to ±0.2 cm, so the radius r = D/2 of the path of the fastest photoelectrons is uncertain by ±0.1 cm.
The table shows the results; the graph shows r² against 1/λ with error bars and the line of best fit. (c = 3.00 × 108 m s−1, e = 1.60 × 10−19 C, me = 9.11 × 10−31 kg)
λ / nm
D / cm
r / cm
1/λ / 106 m−1
r² / cm²
250
11.0
5.50
4.00
30.3
300
9.2
4.60
3.33
21.2
350
7.6
3.80
2.86
14.4
400
6.1
3.05
2.50
9.3
450
4.6
2.30
2.22
5.3
r² against 1/λ with error bars and the line of best fit (drawn to scale).
(a)
(i)
Show that r² = (2me/(e²B²))(hc/λ − Φ), where Φ is the work function of the cathode.
(2)
(b)
(i)
Show that the absolute uncertainty in r² for λ = 350 nm is about 0.8 cm².
(2)
(c)
(i)
Determine, using the graph, a value for the Planck constant.
(2)
(d)
(i)
Determine the work function of the cathode, in eV.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
evB = mev²/r gives r = mev/(eB), so v = eBr/me
✓ 1
Emax = ½mev² = e²B²r²/(2me) = hc/λ − Φ, which rearranges to the given expression
✓ 1
Use of Einstein’s equation is needed.
Part (b)(i)
Fractional uncertainty in r = 0.1/3.8 = 2.6 %; in r² it is twice this, 5.3 %
✓ 1
Δ(r²) = 0.053 × 14.4 = 0.76 cm²
✓ 1
An answer to at least 2 s.f. or full working is required.
Part (c)(i)
Gradient = 14.0 cm² per 106 m−1 = 1.40 × 10−9 m³
✓ 1
Accept 13.5–14.5 cm² per 106 m−1, from two well-separated points on the line.
Allow ECF from (a)(i) and from the gradient. Accept 6.3–6.8 × 10−34 J s.
Part (d)(i)
The line meets the 1/λ axis at 1/λ0 = 1.84 × 106 m−1; Φ = hc/λ0 = 2.3 eV
✓ 1
Accept 2.2–2.4 eV (intercept 1.75–1.95 × 106 m−1). Allow ECF from (c)(i) if the candidate’s value of h is used.
Answers: (b)(i) 0.76 cm² · (c)(i) 6.6 × 10−34 J s · (d)(i) 2.3 eV (the remaining parts are explanations — see the table above)
Syllabus understandingE.2 — the photoelectric effect as evidence of the particle nature of light; Einstein’s explanation using the work function and the maximum kinetic energy of the photoelectrons as given by Emax = hf − Φ where Φ is the work function of the metal; D.3 — the motion of a charged particle in a uniform magnetic field, F = qvB sin θ; Tools 3 — linearising, uncertainty of a squared quantity, gradient and intercept of a graph Command term: Determine
35E-2-02
The photoelectric effect and a charged conductor·E.2 Quantum physics (HL)
Paper 2Medium12 marks
Short answer & extended response7 steps to full marksDetermine
A zinc sphere of radius 2.0 cm is mounted on an insulating stand in a vacuum. The photoelectric properties of zinc were first measured in a photocell: the graph shows the stopping potential Vs against the frequency f of the radiation incident on a zinc cathode.
(c = 3.00 × 108 m s−1, e = 1.60 × 10−19 C, k = 8.99 × 109 N m² C−2)
Stopping potential against frequency for zinc (graph drawn to scale)
(a)
(i)
Determine, using the graph, a value for the Planck constant.
(2)
(ii)
Determine the work function of zinc in eV.
(2)
(iii)
On the graph, draw the line that would be obtained for a metal with a larger work function.
(1)
(b)
The sphere, initially uncharged, is now illuminated with ultraviolet radiation of wavelength 200 nm.
(i)
Determine the maximum kinetic energy, in eV, of the electrons emitted while the sphere is uncharged.
(2)
(ii)
Explain why, after some time, no more electrons leave the sphere, and state the potential of the sphere when this happens.
(2)
(iii)
Calculate the charge on the sphere when emission stops.
(2)
(iv)
Hence determine the number of electrons that have left the sphere.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
Gradient = ΔVs/Δf ≈ 4.1 × 10−15 V s
✓ 1
Accept 4.0–4.3 × 10−15 V s from two well-separated points on the line.
h = e × gradient = 6.6 × 10−34 J s
✓ 1
Accept 6.4–6.9 × 10−34 J s.
Part (a)(ii)
Threshold frequency from the intercept on the f-axis: f0 = 1.04 × 1015 Hz
✓ 1
Accept 1.02–1.06 × 1015 Hz.
Φ = hf0 = 6.9 × 10−19 J = 4.3 eV
✓ 1
Allow ECF from (a)(i). Accept 4.2–4.4 eV. ALT: extrapolating the line to f = 0 gives Vs = −Φ/e.
Part (a)(iii)
A straight line parallel to the given line (same gradient h/e), displaced to higher frequency
✓ 1
Allow ECF from (a)(i). Both features needed.
Part (b)(i)
Photon energy = hc/λ = 9.95 × 10−19 J = 6.22 eV
✓ 1
Emax = 6.22 − 4.3 = 1.9 eV
✓ 1
Allow ECF from (a)(ii).
Part (b)(ii)
Each emitted electron leaves the sphere more positive, so its potential rises and an escaping electron must do work eV against the attraction of the sphere
✓ 1
Emission stops when eV = Emax: V = +1.9 V
✓ 1
Allow ECF from (b)(i). MP2 needs the value.
Part (b)(iii)
V = kQ/r, so Q = Vr/k = 1.92 × 0.020/8.99 × 109
✓ 1
Q = 4.3 × 10−12 C
✓ 1
Allow ECF from (b)(ii). Accept 4.2–4.4 × 10−12 C.
Part (b)(iv)
N = Q/e = 2.7 × 107
✓ 1
Allow ECF from (b)(iii).
Answers: (a)(i) 6.6 × 10−34 J s · (a)(ii) 4.3 eV · (b)(i) 1.9 eV · (b)(ii) +1.9 V · (b)(iii) 4.3 × 10−12 C · (b)(iv) 2.7 × 107(the remaining parts are explanations — see the table above)
Syllabus understandingE.2 (HL) — the photoelectric effect as evidence of the particle nature of light; the threshold frequency; Einstein's explanation, Emax = hf − Φ; D.2 — electric potential Ve = kQ/r and the work done moving a charge Command term: Determine
Short answer & extended response6 steps to full marksDeduce
Neutrons from a research reactor pass through a block of heavy water at 320 K and emerge in thermal equilibrium with it. They are then diffracted by a crystal to reveal the arrangement of its atoms.
(kB = 1.38 × 10−23 J K−1, mass of neutron = 1.675 × 10−27 kg, me = 9.11 × 10−31 kg, h = 6.63 × 10−34 J s, e = 1.60 × 10−19 C)
(a)
(i)
Show that the average kinetic energy of the neutrons is about 6.6 × 10−21 J.
(1)
(ii)
Determine the de Broglie wavelength of a neutron with this kinetic energy.
(2)
(iii)
Explain why these neutrons are suitable for studying the arrangement of atoms in a crystal.
(1)
(b)
(i)
Electrons with the same de Broglie wavelength could also be used. Deduce, without calculating a wavelength, the ratio (kinetic energy of the electron)/(kinetic energy of the neutron).
(2)
(ii)
Hence calculate the potential difference through which the electrons must be accelerated from rest.
(2)
(iii)
Suggest why electrons of this energy are diffracted only by the outermost layers of atoms of a crystal, whereas the neutrons pass through the whole crystal.
p = √(2mEk) = √(2 × 1.675 × 10−27 × 6.62 × 10−21) = 4.71 × 10−24 kg m s−1
✓ 1
Allow ECF from (a)(i).
λ = h/p = 1.4 × 10−10 m
✓ 1
Accept 1.4 × 10−10 m.
Part (a)(iii)
Their wavelength (≈ 10−10 m) is comparable to the spacing of atoms in a crystal, so they are diffracted significantly
✓ 1
Allow ECF from (a)(ii).
Part (b)(i)
The same wavelength means the same momentum (p = h/λ)
✓ 1
Ek = p²/2m, so the ratio = (mass of neutron)/(mass of electron) = 1840
✓ 1
Accept 1.8 × 103. An inverted ratio scores [1].
Part (b)(ii)
Ek = 1839 × 6.62 × 10−21 = 1.22 × 10−17 J
✓ 1
Allow ECF from (a)(i) and (b)(i).
V = Ek/e = 76 V
✓ 1
Accept 75–77 V.
Part (b)(iii)
Electrons are charged and interact strongly (electric force) with the atoms, so they are stopped or scattered near the surface; neutrons are uncharged
✓ 1
Answers: (a)(ii) 1.4 × 10−10 m · (b)(i) 1.8 × 103 · (b)(ii) 76 V (the remaining parts are explanations — see the table above)
Syllabus understandingE.2 (HL) — diffraction of particles as evidence of the wave nature of matter; the de Broglie wavelength λ = h/p; B.1 — that Kelvin temperature is a measure of the average kinetic energy of particles as given by Ek = (3/2)kBT; D.2 — work done on a charge accelerated through a potential difference Command term: Deduce
37E-2-19
The photoelectric effect·E.2 Quantum physics (HL)
Paper 2Medium12 marks
Short answer & extended response8 steps to full marksExplain
An ultraviolet flame detector contains a photocell whose metal cathode has a work function of 4.70 eV. The detector is used in a factory to confirm that a gas burner is alight. (h = 6.63 × 10−34 J s, c = 3.00 × 108 m s−1, e = 1.60 × 10−19 C) (me = 9.11 × 10−31 kg)
(a)
Threshold.
(i)
Calculate the threshold frequency for the cathode.
(1)
(ii)
Sunlight at the Earth's surface contains no wavelengths shorter than about 295 nm. Explain why the detector does not respond to sunlight, however bright it is.
(2)
(b)
Photoelectrons.
(i)
The flame emits ultraviolet radiation of wavelength 220 nm. Determine the maximum kinetic energy, in eV, of the photoelectrons.
(2)
(ii)
Calculate the maximum speed of the photoelectrons.
(1)
(c)
The wave model.
(i)
The intensity of the 220 nm radiation at the cathode is 2.0 × 10−6 W m−2. In the classical wave model, the energy is spread uniformly over the wavefront and an electron can absorb only the energy falling on an area of about 1.0 × 10−19 m² around it. Estimate the time for an electron to absorb enough energy to escape.
(2)
(ii)
The detector actually responds within a few milliseconds of the burner igniting. Explain how the photon model accounts for this.
(2)
(d)
The current.
(i)
The cathode has an area of 1.5 × 10−4 m² and one photon in ten releases an electron. Determine the photoelectric current.
v = √(2Emax/me) = √(2 × 0.95 × 1.60 × 10−19/9.11 × 10−31) = 5.8 × 105 m s−1
✓ 1
ECF from (b)(i).
Part (c)(i)
Power absorbed by one electron = 2.0 × 10−6 × 1.0 × 10−19 = 2.0 × 10−25 W
✓ 1
t = Φ/power = 4.70 × 1.60 × 10−19/2.0 × 10−25 = 3.8 × 106 s (about 44 days)
✓ 1
Work function must be in joules.
Part (c)(ii)
The energy of the radiation is concentrated in photons, each of energy hf (5.65 eV)
✓ 1
A single photon is absorbed by a single electron in one event, so an electron can be emitted as soon as the first photon arrives — no accumulation of energy is needed
✓ 1
Part (d)(i)
Photons per second = (2.0 × 10−6 × 1.5 × 10−4)/9.04 × 10−19 = 3.3 × 108 s−1
✓ 1
I = 0.10 × 3.3 × 108 × 1.60 × 10−19 = 5.3 × 10−12 A
✓ 1
ECF.
Answers: (a)(i) 1.1 × 1015 Hz · (b)(i) 0.95 eV · (b)(ii) 5.8 × 105 m s−1 · (c)(i) 3.8 × 106 s · (d)(i) 5.3 × 10−12 A (the remaining parts are explanations — see the table above)
Syllabus understandingE.2 (HL) — the photoelectric effect as evidence of the particle nature of light; the threshold frequency; Einstein's explanation, Emax = hf − Φ; which features of the photoelectric effect cannot be explained using the classical wave theory of light Command term: Explain
38E-2-20
Compton scattering·E.2 Quantum physics (HL)
Paper 2Hard14 marks
Short answer & extended response9 steps to full marksDetermine
In a hospital gamma camera, a patient is given technetium-99m, which emits gamma photons of energy 140 keV. Some photons are Compton scattered by electrons in the patient's body before reaching the camera. Treat these electrons as free and at rest. (h = 6.63 × 10−34 J s, c = 3.00 × 108 m s−1, e = 1.60 × 10−19 C) (me = 9.11 × 10−31 kg)
(a)
A single scattering.
(i)
Calculate the wavelength of a 140 keV photon.
(1)
(ii)
A 140 keV photon is scattered through 90°. Calculate the wavelength and the energy, in keV, of the scattered photon.
(2)
(iii)
State the kinetic energy given to the electron.
(1)
(b)
The energy window.
(i)
The camera records only photons with energies above 126 keV. Determine the largest angle through which a photon can be scattered and still be recorded.
(3)
(ii)
Explain why the camera is designed to reject scattered photons.
(2)
(c)
The recoiling electron.
(i)
For the photon scattered through 90° in (a)(ii), determine the angle between the direction of motion of the recoiling electron and the direction of the incident photon.
(2)
(d)
Back-scattering.
(i)
A 140 keV photon is scattered back through 180°. Determine the magnitude of the momentum of the recoiling electron.
Δλ = (h/mec)(1 − cos 90°) = 2.43 × 10−12 m, so λf = 8.88 + 2.43 = 11.3 × 10−12 m
✓ 1
E = hc/λf = 1.76 × 10−14 J = 110 keV
✓ 1
ECF; accept 109–111 keV.
Part (a)(iii)
140 − 110 = 30 keV (energy conservation)
✓ 1
ECF.
Part (b)(i)
λf = hc/(126 keV) = 9.87 × 10−12 m, so Δλ = 9.87 − 8.88 = 0.99 × 10−12 m
✓ 1
1 − cos θ = 0.99/2.43 = 0.41, so cos θ = 0.59
✓ 1
θ = 54°
✓ 1
Accept 53–55°. Scattering through a larger angle gives a photon below 126 keV.
Part (b)(ii)
A scattered photon has changed direction, so it appears to come from a point in the body where the source is not
✓ 1
Recording it would blur the image / add background; since scattering lowers the energy, an energy threshold removes most scattered photons
✓ 1
Part (c)(i)
Momentum is conserved in two perpendicular directions: along the incident direction the electron's momentum component is h/λi; perpendicular to it, it is h/λf, opposite to the momentum of the scattered photon
✓ 1
Allow ECF from (a)(i) and (a)(ii).
tan φ = (h/λf)/(h/λi) = λi/λf = 8.88/11.3, so φ = 38° (on the opposite side of the incident direction from the scattered photon)
✓ 1
Accept 37–39°. The inverted ratio (52°) scores [1].
Part (d)(i)
λf = 8.88 + 2 × 2.43 = 13.7 × 10−12 m
✓ 1
pi = h/λi = 7.47 × 10−23 N s; pf = h/λf = 4.83 × 10−23 N s in the opposite direction
✓ 1
Momentum conservation: pe = pi + pf = 1.2 × 10−22 N s
✓ 1
Adding (not subtracting) the photon momenta is the key step; accept 1.2–1.3 × 10−22 N s.
Answers: (a)(i) 8.88 × 10−12 m · (a)(ii) 1.13 × 10−11 m, 110 keV · (a)(iii) 30 keV · (b)(i) 54° · (c)(i) 38° · (d)(i) 1.2 × 10−22 N s (the remaining parts are explanations — see the table above)
Syllabus understandingE.2 (HL) — Compton scattering of light by electrons as additional evidence of the particle nature of light; that photons scatter off electrons with increased wavelength; λf − λi = Δλ = (h/mec)(1 − cos θ); A.2 — conservation of linear momentum; guidance: a quantitative approach to collisions is for two-dimensional situations for higher level students Command term: Determine
39E-2-36
Electrons through a double slit·E.2 Quantum physics (HL)
Paper 2Hard12 marks
Short answer & extended response8 steps to full marksDetermine
Electrons, accelerated from rest through a potential difference of 600 V, pass through two very narrow parallel slits whose centres are 2.5 × 10−7 m apart, cut in a thin membrane. A position-sensitive detector, 0.60 m beyond the slits, records the point of arrival of each electron as a single dot. The beam current is so small that the dots appear one at a time; after a long time they build up the distribution shown in the graph.
(h = 6.63 × 10−34 J s, c = 3.00 × 108 m s−1, e = 1.60 × 10−19 C, me = 9.11 × 10−31 kg)
Distribution of electron arrivals across the detector after a long time (graph drawn to scale)
(a)
(i)
Show that the de Broglie wavelength of the electrons is about 5.0 × 10−11 m.
(2)
(ii)
Determine, using the graph, the separation of adjacent maxima on the detector.
(2)
(iii)
Hence determine the wavelength of the electrons, and comment on your answer.
(2)
(b)
(i)
The beam current at the slits is 1.6 × 10−15 A. Determine the average distance between one electron and the next in the beam.
(3)
(ii)
Explain what your answer to (b)(i) shows about the formation of the pattern.
(2)
(c)
(i)
State the feature of the experiment that shows the particle nature of the electrons.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
p = √(2meV) = √(2 × 9.11 × 10−31 × 1.60 × 10−19 × 600) = 1.32 × 10−23 kg m s−1
✓ 1
λ = h/p = 5.01 × 10−11 m
✓ 1
An answer to at least 3 s.f. must be seen.
Part (a)(ii)
Measure across several spacings, e.g. the maxima at about −0.48 mm and +0.48 mm (0.95 mm apart on the graph) are 8 spacings apart
✓ 1
Measuring one spacing only: [1 max].
s = 0.95/8 = 0.119 mm
✓ 1
Accept 0.115–0.125 mm (0.96/8 = 0.120 mm accepted).
The electrons are about 1.5 km apart, so there is almost never more than one electron between the slits and the detector (0.60 m) at a time
✓ 1
Allow ECF from (b)(i).
The fringes therefore cannot come from different electrons interfering with each other: each single electron behaves as a wave that passes through both slits and interferes with itself
✓ 1
OWTTE. Dependent on MP1.
Part (c)(i)
Each electron arrives at a single point on the detector (one dot, all of its charge at one place), not spread over the pattern
✓ 1
Do not accept "electrons have mass".
Answers: (a)(ii) 0.12 mm · (a)(iii) 5.0 × 10−11 m · (b)(i) 1.5 × 103 m (the remaining parts are explanations — see the table above)
Syllabus understandingE.2 (HL) — diffraction of particles as evidence of the wave nature of matter; that matter exhibits wave–particle duality; the de Broglie wavelength for particles as given by λ = h/p; C.3 — double-slit interference, s = λD/d; D.2 — work done on a charge, W = qV; B.5 — electric current as a flow of charge carriers, I = Δq/ΔtCommand term: Determine
40E-2-42
Hydrogen light on a photocell·E.2 Quantum physics (HL)
Paper 2Easy13 marks
Short answer & extended response8 steps to full marksDetermine
Light from a hydrogen discharge lamp falls on the cathode of a vacuum photocell. The cathode has a work function of 2.30 eV. The glass of the lamp and of the photocell absorbs ultraviolet radiation, so the only light reaching the cathode is in the four visible spectral lines of hydrogen. These are produced by transitions from the levels n = 3, 4, 5 and 6 to the level n = 2. The table gives the energies of these levels.
(h = 6.63 × 10−34 J s, c = 3.00 × 108 m s−1, e = 1.60 × 10−19 C, me = 9.11 × 10−31 kg)
n
2
3
4
5
6
En / eV
−3.40
−1.51
−0.85
−0.54
−0.38
(a)
(i)
Show that the wavelength of the photons emitted in the transition from n = 4 to n = 2 is about 490 nm.
(2)
(ii)
Explain why the red light, from the transition n = 3 to n = 2, releases no photoelectrons.
(1)
(b)
(i)
Determine the stopping potential for the photocell.
(2)
(ii)
A filter that absorbs only the 488 nm line is placed in front of the photocell. State and explain the effect on the stopping potential and on the maximum photoelectric current.
(2)
(c)
(i)
Calculate the maximum speed of the photoelectrons.
(2)
(ii)
Determine the de Broglie wavelength of these photoelectrons.
(1)
(d)
(i)
The cathode is replaced by one with a work function of 2.90 eV. Deduce which of the four lines release photoelectrons from the new cathode, and determine the new stopping potential.
Photon energies: 1.89 eV, 2.55 eV, 2.86 eV and 3.02 eV for n = 3, 4, 5 and 6 → 2
✓ 1
Allow ECF from (a) and (b)(i).
Only the n = 6 → 2 line (3.02 eV) exceeds 2.90 eV; the n = 5 → 2 line (2.86 eV) does not
✓ 1
Vs = 3.02 − 2.90 = 0.12 V
✓ 1
Answers: (a)(i) 488 nm · (b)(i) 0.72 V · (c)(i) 5.03 × 105 m s−1 · (c)(ii) 1.4 × 10−9 m · (d)(i) only the 412 nm line; 0.12 V (the remaining parts are explanations — see the table above)
Syllabus understandingE.2 — the photoelectric effect as evidence of the particle nature of light; that photons of a certain frequency, known as the threshold frequency, are required to release photoelectrons from the metal; Einstein’s explanation using the work function and the maximum kinetic energy of the photoelectrons as given by Emax = hf − Φ where Φ is the work function of the metal; the de Broglie wavelength for particles as given by λ = h/p; E.1 — photons emitted in transitions between discrete energy levels, E = hf; A.3 — Ek = ½mv² Command term: Determine
41E-2-43
A low-voltage electron microscope·E.2 Quantum physics (HL)
Paper 2Easy10 marks
Short answer & extended response7 steps to full marksDetermine
In a low-voltage electron microscope, electrons are accelerated from rest through a potential difference of 5.00 kV and then pass through a thin specimen. Detail in the specimen can be seen only if it is not much smaller than the wavelength of the waves used, because waves diffract around very small objects.
(h = 6.63 × 10−34 J s, c = 3.00 × 108 m s−1, e = 1.60 × 10−19 C, me = 9.11 × 10−31 kg)
(a)
(i)
Calculate, in J, the kinetic energy of an electron after it has been accelerated.
(1)
(ii)
Show that the speed of the electrons is about 4 × 107 m s−1, and that at this speed the Lorentz factor is close to 1.
(2)
(b)
(i)
Determine the de Broglie wavelength of the electrons.
(2)
(ii)
A light microscope uses light of wavelength 500 nm. Explain, with a calculation, why the electron microscope can show much finer detail.
(2)
(c)
(i)
The accelerating potential difference is increased to 20.0 kV. Deduce the new de Broglie wavelength of the electrons. Ignore relativistic effects.
(1)
(d)
(i)
In a magnetic lens, the electrons accelerated through 5.00 kV move at right angles to a uniform magnetic field of flux density 2.0 mT. Calculate the radius of their circular path.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
Ek = eV = 1.60 × 10−19 × 5.00 × 103 = 8.00 × 10−16 J
✓ 1
Part (a)(ii)
v = √(2Ek/me) = √(2 × 8.00 × 10−16/9.11 × 10−31) = 4.19 × 107 m s−1
✓ 1
Allow ECF from (a)(i). An answer to at least 2 s.f. or full substitution is required.
γ = 1/√(1 − (4.19 × 107/3.00 × 108)²) = 1.01, so relativistic effects are about 1 %
✓ 1
Accept v/c = 0.14 with γ ≈ 1.01.
Part (b)(i)
p = mev = 9.11 × 10−31 × 4.19 × 107 = 3.82 × 10−23 kg m s−1
✓ 1
Allow ECF from (a)(ii). ALT: p = √(2meEk).
λ = h/p = 1.74 × 10−11 m
✓ 1
Part (b)(ii)
Ratio of wavelengths = 500 × 10−9/1.74 × 10−11 = 2.9 × 104
✓ 1
Allow ECF from (b)(i).
Detail much smaller than the wavelength cannot be seen because the waves diffract around it; the electron wavelength is about 3 × 104 times shorter, so detail about 3 × 104 times finer can be seen
✓ 1
Reference to diffraction or to the size of the detail relative to the wavelength is needed.
Part (c)(i)
p = √(2meeV) ∝ √V, so λ ∝ 1/√V: multiplying V by 4 halves λ, giving 8.7 × 10−12 m
✓ 1
Allow ECF from (b)(i).
Part (d)(i)
evB = mev²/r, so r = mev/(eB)
✓ 1
r = 9.11 × 10−31 × 4.19 × 107/(1.60 × 10−19 × 2.0 × 10−3) = 0.12 m
✓ 1
Allow ECF from (a)(ii). Accept 0.119 m.
Answers: (a)(i) 8.00 × 10−16 J · (a)(ii) 4.19 × 107 m s−1; γ = 1.01 · (b)(i) 1.74 × 10−11 m · (b)(ii) 2.9 × 104 · (c)(i) 8.7 × 10−12 m · (d)(i) 0.12 m (the remaining parts are explanations — see the table above)
Syllabus understandingE.2 — that matter exhibits wave–particle duality; the de Broglie wavelength for particles as given by λ = h/p; diffraction of particles as evidence of the wave nature of matter; D.2 (HL) — W = qΔVe; D.3 — the motion of a charged particle in a uniform magnetic field; A.5 — γ = 1/√(1 − v²/c²) Command term: Determine
42E-2-44
Counting photons from a star·E.2 Quantum physics (HL)
Paper 2Medium11 marks
Short answer & extended response7 steps to full marksDetermine
A photomultiplier on a telescope detects light from a star of luminosity 9.6 × 1026 W at a distance of 25 pc. A filter transmits only a narrow band of wavelengths around 450 nm; 2.0 % of the luminosity of the star is in this band. The telescope collects light over an area of 0.80 m². The photocathode of the photomultiplier has a work function of 1.90 eV, and on average one in four of the photons that reach it releases a photoelectron. Ignore absorption in the atmosphere and in the telescope.
(h = 6.63 × 10−34 J s, c = 3.00 × 108 m s−1, e = 1.60 × 10−19 C, 1 pc = 3.09 × 1016 m)
(a)
(i)
Show that the apparent brightness of the star is about 1.3 × 10−10 W m−2.
(2)
(ii)
Determine the number of photons in the 450 nm band that enter the telescope each second.
(3)
(b)
(i)
Determine the current of photoelectrons leaving the photocathode.
(2)
(ii)
Calculate, in eV, the maximum kinetic energy of the photoelectrons.
(1)
(c)
(i)
The telescope is pointed at a second star with the same luminosity and spectrum as the first, at a distance of 50 pc. State and explain the effect on the photoelectric current and on the maximum kinetic energy of the photoelectrons.
(2)
(ii)
For the first star, estimate the average time between the emission of successive photoelectrons.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
d = 25 × 3.09 × 1016 = 7.73 × 1017 m
✓ 1
b = L/4πd² = 9.6 × 1026/(4π × (7.73 × 1017)²) = 1.28 × 10−10 W m−2
✓ 1
Full substitution or an answer to at least 3 s.f. is required.
Part (a)(ii)
Power in the band = 0.020 × 1.3 × 10−10 × 0.80 = 2.1 × 10−12 W
Number per second = 2.08 × 10−12/4.42 × 10−19 = 4.7 × 106 s−1
✓ 1
Accept 4.6–4.7 × 106 s−1.
Part (b)(i)
Photoelectrons per second = 0.25 × 4.71 × 106 = 1.18 × 106 s−1
✓ 1
Allow ECF from (a)(ii).
I = 1.18 × 106 × 1.60 × 10−19 = 1.9 × 10−13 A
✓ 1
Part (b)(ii)
Photon energy = 4.42 × 10−19 J = 2.76 eV, so Emax = 2.76 − 1.90 = 0.86 eV
✓ 1
Allow ECF from (a)(ii).
Part (c)(i)
The current falls to a quarter (4.7 × 10−14 A): b ∝ 1/d², so a quarter as many photons, and so photoelectrons, arrive each second
✓ 1
Allow ECF from (b)(i).
Emax is unchanged (0.86 eV): each photon still has energy hc/λ, and one photon gives its energy to one electron
✓ 1
Allow ECF from (b)(ii).
Part (c)(ii)
t = 1/1.18 × 106 = 8.5 × 10−7 s
✓ 1
Allow ECF from (b)(i). Accept 8.4–8.7 × 10−7 s.
Answers: (a)(i) 1.28 × 10−10 W m−2 · (a)(ii) 4.7 × 106 s−1 · (b)(i) 1.9 × 10−13 A · (b)(ii) 0.86 eV · (c)(i) a quarter; unchanged · (c)(ii) 8.5 × 10−7 s (the remaining parts are explanations — see the table above)
Syllabus understandingE.2 — the photoelectric effect as evidence of the particle nature of light; Einstein’s explanation using the work function and the maximum kinetic energy of the photoelectrons as given by Emax = hf − Φ where Φ is the work function of the metal; B.1 — apparent brightness b = L/4πd²; E.5 — astronomical distances in parsecs; B.5 — electric current I = Δq/ΔtCommand term: Determine
43E-2-45
Slowing atoms with laser light·E.2 Quantum physics (HL)
Paper 2Medium13 marks
Short answer & extended response8 steps to full marksDetermine
In an atomic-beam experiment, sodium atoms leave an oven travelling at 600 m s−1. A laser beam of wavelength 589 nm is directed against the atoms, opposite to their direction of motion. An atom absorbs a photon from the laser beam and later re-emits a photon of the same wavelength in a random direction; it can then absorb another photon from the beam. The mass of a sodium atom is 23 u.
Show that the momentum of a photon of the laser light is about 1.1 × 10−27 kg m s−1.
(1)
(ii)
Determine the change in speed of an atom when it absorbs one photon.
(2)
(b)
(i)
Explain why the atom slows down even though it re-emits a photon after each absorption.
(2)
(ii)
Determine the number of photons that the atom must absorb to bring it almost to rest.
(1)
(c)
(i)
Each absorption and re-emission takes at least 32 ns. Determine the shortest time in which an atom can be brought to rest and the distance it travels in this time. Assume that the deceleration is uniform.
(3)
(d)
After cooling, the atoms move with speeds of about 0.30 m s−1.
(i)
Determine the de Broglie wavelength of an atom moving at 0.30 m s−1, and compare it with the diameter of a sodium atom, about 3 × 10−10 m.
(2)
(ii)
Estimate the temperature of the cooled atoms, and state one assumption you make.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
p = h/λ = 6.63 × 10−34/589 × 10−9 = 1.13 × 10−27 kg m s−1
✓ 1
Full substitution or an answer to at least 3 s.f. is required.
Part (a)(ii)
m = 23 × 1.66 × 10−27 = 3.82 × 10−26 kg; momentum is conserved, so the momentum of the atom decreases by h/λ
Allow ECF from (a)(i). Accept 2.9–3.0 × 10−2 m s−1.
Part (b)(i)
Every absorbed photon comes from the same direction, opposite to the motion of the atom, so each absorption reduces the forward momentum of the atom and the changes add up
✓ 1
The photons are re-emitted in random directions, so over many cycles the momentum changes due to emission average to zero
✓ 1
Do not accept “the emitted photon has less momentum”.
Part (b)(ii)
N = 600/2.95 × 10−2 = 2.0 × 104
✓ 1
Allow ECF from (a)(ii).
Part (c)(i)
t = 2.04 × 104 × 32 × 10−9 = 6.5 × 10−4 s
✓ 1
Allow ECF from (b)(ii). Accept 6.4–6.5 × 10−4 s.
Average speed = (600 + 0)/2 = 300 m s−1 (or deceleration = 2.95 × 10−2/32 ns = 9.2 × 105 m s−2)
About 200 times the diameter of the atom, so the wave nature of the cold atoms is significant (they can be diffracted by structures much larger than an atom)
✓ 1
A comparison and a conclusion are needed.
Part (d)(ii)
½mv² = (3/2)kBT, so T = mv²/(3kB) = 3.82 × 10−26 × 0.30²/(3 × 1.38 × 10−23) = 8.3 × 10−5 K
✓ 1
Allow ECF from (a)(ii) for the mass.
Assumption: 0.30 m s−1 is the root mean square speed of the atoms (or: the cooled atoms behave as an ideal gas)
✓ 1
Answers: (a)(i) 1.13 × 10−27 kg m s−1 · (a)(ii) 2.95 × 10−2 m s−1 · (b)(ii) 2.0 × 104 · (c)(i) 6.5 × 10−4 s; 0.20 m · (d)(i) 5.8 × 10−8 m · (d)(ii) 8.3 × 10−5 K (the remaining parts are explanations — see the table above)
Syllabus understandingE.2 — that matter exhibits wave–particle duality; the de Broglie wavelength for particles as given by λ = h/p; applied to photons, p = h/λ; A.2 — conservation of linear momentum; A.1 — equations of motion for uniformly accelerated motion; B.1 — Ek = (3/2)kBTCommand term: Determine
44E-2-46
Compton scattering in a gamma-ray detector·E.2 Quantum physics (HL)
Paper 2Hard14 marks
Short answer & extended response9 steps to full marksDetermine
A gamma-ray detector contains a large crystal. When a gamma photon enters the crystal, one of two things usually happens: the photon is absorbed completely by an electron (photoelectric absorption), or the photon is Compton scattered by an electron and the scattered photon then escapes from the crystal. In both cases the detector measures the kinetic energy given to the electron. Treat the electrons as free and initially at rest, and ignore their binding energy.
The graph shows the number of events per unit energy against the kinetic energy of the electron, for gamma photons of a single energy from a radioactive source.
(h = 6.63 × 10−34 J s, c = 3.00 × 108 m s−1, e = 1.60 × 10−19 C, λC = h/mec = 2.43 × 10−12 m)
Number of events per unit energy against the kinetic energy given to the electron (schematic, drawn to scale).
(a)
(i)
State, with a reason, the energy of the gamma photons emitted by the source.
(1)
(ii)
Show that the wavelength of these photons is about 1.5 × 10−12 m.
(2)
(b)
(i)
Explain why, in a Compton event, the electron receives the greatest kinetic energy when the photon is scattered through 180°.
(2)
(ii)
Determine, in keV, the greatest kinetic energy that an electron can receive in a Compton event in this detector.
(3)
(iii)
Outline how the graph supports your answer to (b)(ii).
(1)
(c)
(i)
Show that the greatest kinetic energy given to an electron in a Compton event is Kmax = E × 2λC/(λ + 2λC), where E and λ are the energy and the wavelength of the incident photon.
(2)
(ii)
For gamma photons of very high energy, the difference E − Kmax between the photon energy and the greatest Compton kinetic energy approaches a constant value. Deduce this value in keV.
(2)
(d)
(i)
Every gamma photon from this source has the same energy. State what this suggests about the nucleus that emits them.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
835 keV: the narrow peak at 835 keV is due to photoelectric absorption, in which the electron receives all the energy of the photon
Full substitution or an answer to at least 3 s.f. is required.
Part (b)(i)
Δλ = λC(1 − cos θ) is greatest when cos θ = −1, i.e. θ = 180°, where Δλ = 2λC
✓ 1
The scattered photon then has its longest wavelength and so its smallest energy (E = hc/λ); by conservation of energy the electron receives the most energy
The nucleus has discrete energy levels: each gamma photon is emitted in a transition between two levels, and its energy equals the difference between them
✓ 1
Do not accept “the nucleus is stable”.
Answers: (a)(i) 835 keV · (a)(ii) 1.49 × 10−12 m · (b)(ii) 639 keV · (c)(ii) 256 keV (the remaining parts are explanations — see the table above)
Syllabus understandingE.2 — Compton scattering of light by electrons as additional evidence of the particle nature of light; that photons scatter off electrons with increased wavelength; the shift in photon wavelength after scattering off an electron as given by λf − λi = Δλ = (h/mec)(1 − cos θ); the photoelectric effect as evidence of the particle nature of light; E.3 (HL) — gamma spectra as evidence for discrete nuclear energy levels; E.1 — E = hfCommand term: Determine
45E-2-64
A solar sail·E.2 Quantum physics (HL)
Paper 2Hard19 marks
Short answer & extended response10 steps to full marksDetermine
A solar sail is a large, very thin sheet that is accelerated by the momentum of sunlight. A sail of area 1600 m² has a total mass (sail and payload) of 12 kg. Sunlight strikes the sail at normal incidence and is completely reflected.
(L☉ = 3.85 × 1026 W, M☉ = 1.99 × 1030 kg, 1 AU = 1.50 × 1011 m, G = 6.67 × 10−11 N m² kg−2, h = 6.63 × 10−34 J s, c = 3.00 × 108 m s−1)
Acceleration of the sail due to sunlight against 1/d² (graph drawn to scale)
(a)
(i)
Show that the momentum of a photon of wavelength 550 nm is about 1.2 × 10−27 kg m s−1.
(1)
(ii)
Estimate the number of photons striking the sail each second when it is 1.00 AU from the Sun.
(2)
(iii)
State one assumption made in your estimate in (a)(ii).
(1)
(iv)
Show that the force exerted by the sunlight on the sail is F = 2bA/c, where b is the intensity of the sunlight and A is the area of the sail.
(2)
(v)
Determine the acceleration of the sail at 1.00 AU due to the sunlight.
(2)
(b)
(i)
Show that the ratio of the force due to sunlight to the gravitational force of the Sun on the sail does not depend on the distance d of the sail from the Sun.
(2)
(ii)
Calculate this ratio for the sail.
(1)
(iii)
Determine the greatest mass per unit area that such a sail (including its payload) could have if sunlight is to be able to balance the Sun's gravitational force on it.
(2)
(c)
(i)
During a mission the acceleration a of the sail due to sunlight alone is measured at different distances d from the Sun. The graph shows a against 1/d². Determine the gradient of the graph, giving its unit.
(2)
(ii)
Hence determine the luminosity of the Sun.
(2)
(d)
(i)
Over time the surface of the sail degrades so that it absorbs 10 % of the incident photons and reflects the rest. Determine the force on the sail at 1.00 AU.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
p = h/λ = 6.63 × 10−34/550 × 10−9 = 1.21 × 10−27 kg m s−1
✓ 1
Must see 1.21 × 10−27 or the full substitution.
Part (a)(ii)
b = L/4πd² = 1.36 × 103 W m−2; power on the sail = 2.18 × 106 W
An absorbed photon gives momentum p, a reflected one 2p: F = (0.90 × 2 + 0.10 × 1)bA/c = 1.9bA/c
✓ 1
Allow ECF from (a)(iv).
F = 1.9 × 1.36 × 103 × 1600/3.00 × 108 = 1.4 × 10−2 N
✓ 1
A force of 0.90 × the original (1.3 × 10−2 N) scores [1].
Answers: (a)(ii) 6.0 × 1024 s−1 · (a)(v) 1.2 × 10−3 m s−2 · (b)(ii) 0.21 · (b)(iii) 1.5 × 10−3 kg m−2 · (c)(i) 2.7 × 1019 m³ s−2 · (c)(ii) 3.9 × 1026 W · (d)(i) 1.4 × 10−2 N (the remaining parts are explanations — see the table above)
Syllabus understandingE.2 (HL) — photons; the momentum of a photon, p = h/λ; A.2 — force as the rate of change of momentum; D.1 — Newton's law of gravitation; B.1 — luminosity and apparent brightness, b = L/4πd²; Tools 3 — gradient of a graph with its unit Command term: Determine
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